While designing buildings and structures, engineers calculate the amount a beam will bend, or deflect under load. The same formulas can be used to estimate how much a V-slot profile will bend, which could be an important consideration for 3D printers and other builds.

Builder

Thanks a lot for this resource. This is the kind that really helps new comers like me

A few questions for experienced makers and engineers around:
* As far as I understand, drawings represent "fixed" beams better than simply supported ones ?
* Aren't values labels inverted in the table for "standard" profiles, and shouldn't the axis be inverted or the cross-section rotated 90° for the C-Beam (also in the table) ?
* How should the calculated values be used in cases where several profiles are combined to form a single beam ? Is it just a matter of adding the moments of inertia of the individual profiles ? Surely, it must be much more complex ! But is it a tolerable approximation ?

It's been a few days that I'm thinking of assembling profiles in "II", "H", or "U" to build a super rigid 1500mm Y gantry for a 2,2KW/5Kg spindle so any further info will be very welcome !

Bests

PS: I've built a simple spreadsheet based on the data in the PDF that I'm attaching here for everyone to use, abuse and comment

Staff MemberModeratorBuilder

@Mark Carew No problem. I should also add that the gantry would be considered a beam with fixed ends. The formula is the same except there is no 5. So, FL^3/384EI.

A fixed end beam has less deflection than a simply supported.
That said. This formula is for a uniform load across the fixed end beam. The z carriage does have a distribution width but we should consider the fixed end with point loading formula which is: FL^3/192EI.
The results are not as rosey as the uniform distribution, but they are much closer to what the deflection will be when the carriage is in the center.

Also, I'm getting different values for the area moments of inertia using solidworks. For instance, a 20x20 is 6291.72 (mm^4)

Staff MemberModeratorResident BuilderBuilder

I should also add that the gantry would be considered a beam with fixed ends.

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I have to differ here. Fixity implies infinite rigidity at the connection. The plates (and other frame parts) we bolt into have some level of flexure and thus will allow some level of rotation at the ends whether it is perceptible or not and thus do not meet the requirements for claiming fixity. The true deflection can only be determined through a fairly complex analysis which involves determining the actual moments at the ends and at the center of the beam and then running them all through the formula at the bottom of the chart below.

As such an analysis is well beyond the average person (or at least the average person without some substantial software) it is always best to assume the beam is simply supported as this is the more conservative approach. It is generally better to have less deflection than estimated than more.

Staff MemberModeratorBuilder

Fixture type doesn't imply infinite ridgidity at all. It does imply how the beam will react at its fixture and can tell us what effect it will have on the fixture if we choose the right fixture to calculate to. The reaction of the fixture is to be analyzed "separately" based on this data. Choosing the correct fixture type initially also gives us a clear understanding of what is happening throughout the beam itself and how much deflection occurs when a load is x distance away from an end. This is the whole reason for the different types of supports. The stresses throughout a beam freely supported differ greatly from those that are fixed nearer the ends. With a simply supported beam the deflection based on location is on an arc. With a fixed beam the deflection based on location is on a long bell curve with less deflection at the ends compared to the SS model. This translates to more force downwards through the fixture throughout the loading range where as a simply supported system would lose vertical force due to intro of a force vector (because of the arc. More below).
A simply supported calculation gives us nothing useful for what is actually happening at the fixture in order to assist with determining what's going on through the rest of the system. Hence no moments in the calc. IF one were to use the angle of deflection at the support point based on the beams bend diameter and the angle of tangent contact they would be extremely off on calculations further down to other components. With a fixed beam there is more downward force regardless of location because of the resistance to tension and compression that the fixture itself provides to the top and bottom of the beam....hence the bell curve.

The flexure of the plate (or effect of the beams moment) that you mentioned is the vital part of this. It's what allows one to go further with calculating what is happening after the beam and into the plates. The difference between true deflection and a reasonable calculation is not within reach. You just have to use the right methods (calculations) to get to a reasonable number. Also, this is not out of reach for anyone including the laymen. Why discourage them?
And no. It's not always best to run it through the calculation of a simply supported beam because it's most conservative. Using an additional factor of 4 is about as least conservative as you can get. It's downright wasteful. Just considering the effect that has on the performance of a machine is a joke.

For any newbs reading this. Don't design your machine for no deflection. It will never perform as expected (if it performs at all) and will cost a fortune.

Sorry Rick. I do this everyday. As a mech designer and assistant to multiple mech engineers I'd be out of a job if I couldn't do this sorta legwork.

Staff MemberModeratorResident BuilderBuilder

Sorry Rick. I do this everyday. As a mech designer and assistant to multiple mech engineers I'd be out of a job if I couldn't do this sorta legwork.

Click to expand...

Sorry Joe, as a licensed structural engineer for over 20 years it was my job to explain to the "designers" when they were off base and your missive is not even close. The beam charts are based on a purist mentality and fixed supports are considered absolutely fixed with no possible rotation at the joint. Such fixity is not possible in a frame system like a gantry frame, especially not in one where a relatively stiff beam is connected to considerably weaker side plates. While the side plates will reduce the deflection a small amount, the result won't be significantly different than designing it as a simply supported beam.

Builder

Well, men, seems like I'll have to make my own assumption on this point so I've added fixed end uniform and centered load calculations to the spreadsheet (updated).

Still have to figure what could be a simple and informative/useful way to add "load at any location"... Is there any more significant distance(s) than the usual suspects (center, end, half way) ?

Also, can I add, respectively, X and Y components of area moment of inertia to calculate deflection of an assembled dual, triple, "L", "I", "H", "C" or "U" beam ?

EDIT: This would mean that two beams made of, say, 2 vertical 20x60 profiles and one 20x20, the first assembled in "|-|" configuration, and the other in "|_|, would have the same stiffness and experience the same deflection under load ?!

Hope someone will light my candle...

Bests

PS: @Mark Carew, the actual files extensions for ODF documents are '.odt' for texts and '.ods' for sheets. I should have been more explicit, sorry for that :/

Builder

@Mark Carew No problem. I should also add that the gantry would be considered a beam with fixed ends. The formula is the same except there is no 5. So, FL^3/384EI.

A fixed end beam has less deflection than a simply supported.
That said. This formula is for a uniform load across the fixed end beam. The z carriage does have a distribution width but we should consider the fixed end with point loading formula which is: FL^3/192EI.
The results are not as rosey as the uniform distribution, but they are much closer to what the deflection will be when the carriage is in the center.

Also, I'm getting different values for the area moments of inertia using solidworks. For instance, a 20x20 is 6291.72 (mm^4)

Joe

Can't delete this smiley

Click to expand...

I double checked the moment of inertia in Solidworks and they seem to be right. I did it by highlighting the end face and clicking on section properties. I converted the values in the table to metres^4 so you don't need to worry about unit conversions in the calculations.

Builder

Regarding the discussion of fixed versus simple supports, I would imagine a bolted joint is somewhere in between, since it is not completely rigid, nor is it free to pivot. But I think the main point here is that these calculations are over simplified anyway. They are useful for making relative comparisons between two different beams, but they are not going to give incredibly accurate predictions of actual deflection values on your build. But that's okay, they can still help to inform your design choices.

Staff MemberModeratorResident BuilderBuilder

I'd love to have the time to do a video on this and demonstrate the accuracy of the math to real world results just to smite you

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Oh, please do. But in the mean time let's just go with a little simple computerized structural analysis.

Based on a double 20x60 V-slot beam spanning 1 meter with a 100 lb load at the center:
Deflection of a simply supported beam = .018"
Deflection of a simple beam with fixed ends = .004"
Deflection of the same beam when analyzed as part of a frame with 1/4" x 6" end plates = .016"

.016" would be closer to which.... .018" or .004"?

I stand by my original position. As for all that drivel you posted from the internet, I read most of that nearly 30 years ago. The difference being I actually took courses to explain what it all meant. It's not as straightforward as you think.

Builder

Arguing engineers are hilarious yet incredibly educational. Long may it continue!

FWIW, I'd like to see an FEA breakdown of the deflections on various structure types arranged into a C shape (ie. a gantry platform and its support pillars with variable corner connections)- from aluminum extrusions through solid 50mm square 1045 bar to triangular-cross-braced quad tubes (like a construction crane) and various half-way permutations (aluminum sandwiched inside high tensile steel? Solid steel bar with cross-drilled circular holes?). It wouldn't necessarily translate to everyone's exact intended use, but it might be a good starting point for comparisons.

Plus, obviously the "best" materials and geometries as they apply to mass and torsional and flexural rigidities may have to be tempered by the needs of vibration damping or other practical concerns. I wonder when we'll start filling 4040 V-Slot with concrete? But yeah, while V-Slot's modularity embraces the precepts of rapid machine design- at least as it applies to hobby-level tooling- it would be equally interesting to see how it stacks up against other, perhaps slower, possibilities.

(PS. I never left, I've been lurking whilst working on actually creating real things other than tooling! Got a LinuxCNC V-Slot machine build planned to start this coming spring, but I just bought a bench lathe, so that's gonna be fun in the meantime!)